Silicon tetrachloride ( $S i C l_{4}$ $SiCl_4$ ) can be prepared by heating $S i$ $Si$ in chlorine gas: $S i (s) + 2 C l_{2} (g) \to S i C l_{4} (l)$ $Si (s) + 2Cl_2(g) -> SiCl_4(l)$ . In one reaction, 0.507 mole of $S i C l_{4}$ $SiCl_4$ is produced. How many moles of molecular chlorine were used in the reaction?

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Silicon tetrachloride ( $S i C l_{4}$ $SiCl_4$ ) can be prepared by heating $S i$ $Si$ in chlorine gas: $S i (s) + 2 C l_{2} (g) \to S i C l_{4} (l)$ $Si (s) + 2Cl_2(g) -> SiCl_4(l)$ . In one reaction, 0.507 mole of $S i C l_{4}$ $SiCl_4$ is produced. How many moles of molecular chlorine were used in the reaction?

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Answer 1 · 2016-08-28T08:40:19

Stoichiometry demands at least $1.14$ $1.14$ $m o l$ $mol$ of $C l_{2}$ $Cl_2$ gas.

Explanation:

The stoichiometric equation, which you have correctly written, clearly and explicitly shows that $1$ $1$ $m o l$ $mol$ of silicon (approx. $28 \cdot g$ $28*g$ ) reacts with $2 \cdot moles$ $2*"moles"$ of chlorine gas (approx. $71 \cdot g$ $71*g$ ) to give $1$ $1$ $m o l$ $mol$ of $S i C l_{4}$ $SiCl_4$ .

The equation establishes both the molar and mass balance.

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Silicon tetrachloride (SiCl4SiCl_4) can be prepared by heating SiSi in chlorine gas: Si(s)+2Cl2(g)→SiCl4(l)Si (s) + 2Cl_2(g) -> SiCl_4(l). In one reaction, 0.507 mole of SiCl4SiCl_4 is produced. How many moles of molecular chlorine were used in the reaction?

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